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a: \(\left(3x-1\right)^2-\left(x+3\right)^3=\left(2-x\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow9x^2-6x+1-x^3-9x^2-27x-27=8-x^3\)
\(\Leftrightarrow-x^3-33x-26-8+x^3=0\)
=>-33x=34
hay x=-34/33
b: \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow x^4-1-x^4+2x^2-1=2\)
\(\Leftrightarrow2x^2=4\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: \(x^2-2\sqrt{3}x+3=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)^2=0\)
hay \(x=\sqrt{3}\)
d: \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-x+\sqrt{2}\right)=0\)
\(\Leftrightarrow x-\sqrt{2}=0\)
hay \(x=\sqrt{2}\)
hẳng đẳng thức tề
(a+b)^2= a^2+2ab+b^2
(a+b)^3= a^3+3a^2b+3ab^2+b^3
a^2-b^2= (a+b)(a-b)
a,\(\left(-\frac{1}{2}x+\frac{1}{4}y^2\right)^2=\left(-\frac{1}{2}x\right)^2+2\left(-\frac{1}{2}x\right).\left(\frac{1}{4}y^2\right)+\left(\frac{1}{4}y^2\right)^2\)
\(=\frac{1}{4}x^2-\frac{1}{4}xy^2+\frac{1}{16}y^4\)
b,\(\left(x+3xy\right)^3=x^3+3.x^2.3xy+3.x.\left(3xy\right)^2+\left(3xy\right)^3\)
\(=x^3+9x^3y+27x^3y^2+27x^3y^3\)
c, \(\left(-2\sqrt{2}+\sqrt{3}\right)^2-\left(\sqrt{3}+3\sqrt{2}\right)^2\)
\(=\left(-2\sqrt{2}\right)^2+2.\left(-2\sqrt{2}\right).\sqrt{3}+\sqrt{3}^2-\left[\sqrt{3}^2+2.3\sqrt{2}.\sqrt{3}+\left(3\sqrt{2}\right)^2\right]\)
\(=4.2-4.\sqrt{6}+3-3-6\sqrt{6}-9.2\)
\(=-10-10\sqrt{6}\)
\(VT\le\sqrt{2\left(1+2x+1+2y\right)}=2\sqrt{1+x+y}\)
\(VT\le2\sqrt{1+\sqrt{2\left(x^2+y^2\right)}}=2\sqrt{3}\)
Dấu "=" xảy ra khi \(x=y=1\)
b) ta có: \(\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x+y\right)^2\ge\left(x+y\right)^2\)
\(\Leftrightarrow2x^2+2y^2\ge\left(x+y\right)^2\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
- Thay \(x^2+y^2=1\)
\(\Rightarrow\)\(2\ge\left(x+y\right)^2\)
\(\Leftrightarrow\sqrt{\left(x+y\right)^2}\le\sqrt{2}\)
\(\Leftrightarrow\left|x+y\right|\le\sqrt{2}\)
\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)
- Áp dụng bđt: \(a^2+b^2+c^2\ge ab+bc+ac\)
có: \(a^4+b^4+c^4\ge a^2b^2+b^2c^2+a^2c^2\) (1)
- Áp dụng tiếp bđt trên
có: \(a^2b^2+b^2c^2+a^2c^2\ge a^2bc+ab^2c+c^2ab\) (2)
\(\Leftrightarrow\)\(a^2b^2+b^2c^2+a^2c^2\ge abc\left(a+b+c\right)\) (3)
(1),(2),(3)\(\Rightarrow\) \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
Áp dụng bđt AM-GM:
\(x^2+\dfrac{1}{x}\ge2\sqrt{x}\)
\(y^2+\dfrac{1}{y}\ge2\sqrt{y}\)
Cộng theo vế: \(VT=x^2+y^2+\dfrac{1}{x}+\dfrac{1}{y}\ge2\left(\sqrt{x}+\sqrt{y}\right)=VP\)
\("="\Leftrightarrow x=y=1\)
\(=\left(x\sqrt{3}\right)^3-3\cdot\left(x\sqrt{3}\right)^2\cdot2+3\cdot x\sqrt{3}\cdot2^2-2^3\)
\(=\left(x\sqrt{3}-2\right)^3\)